Because $\kappa_0<\kappa_1=\operatorname{crit}(j_1)$, the definition of the <critical point of an elementary embedding> immediately gives
$$
j_1(\kappa_0)=\kappa_0.
$$

To evaluate $j_0(\kappa_1)$, first use the regularity of the measurable cardinal $\kappa_1$. Every function $f:\kappa_0\to\kappa_1$ has bounded range, so every ordinal below $j_0(\kappa_1)$ lies below $j_0(\beta)$ for some $\beta<\kappa_1$. Hence
$$
j_0(\kappa_1)=\sup_{\beta<\kappa_1}j_0(\beta).
$$
The <measurable cardinal is a strong limit cardinal>[strong-limit property] of $\kappa_1$ gives $\beta^{\kappa_0}<\kappa_1$ for every $\beta<\kappa_1$. There are therefore fewer than $\kappa_1$ functions $\kappa_0\to\beta$, which implies $j_0(\beta)<\kappa_1$. On the other hand $j_0(\beta)\geq\beta$. Taking suprema yields
$$
\boxed{j_0(\kappa_1)=\kappa_1,
\qquad j_1(\kappa_0)=\kappa_0,}
$$
the <two measurable cardinals under an ultrapower embedding> formula.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 116 / 2 / e / Solution

Let $\kappa_0<\kappa_1<\lambda$, where both $\kappa_i$ are measurable and $\lambda$ is inaccessible, and let $j_i:V_\lambda\to M_i$ be ultrapower embeddings by measures on $\kappa_i$. Then
$$
j_0(\kappa_1)=\kappa_1,
\qquad
j_1(\kappa_0)=\kappa_0.
$$
The second equality follows from the critical point. For the first, regularity gives continuity of $j_0$ at $\kappa_1$, while strong-limitness bounds $j_0(\beta)<\kappa_1$ for every $\beta<\kappa_1$.

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